Matrix multiplication in "linear" categories (Exercise 23.1 in Conceptual Mathematics) This is a follow-up to my question here. In Lawvere's Conceptual Mathematics, linear categories (apparently called additive categories elsewhere) are defined as seen in this paragraph:
Linear category definition
Lawvere proceeds to define a 'product' of two matrices in a linear category as follows:
Matrix product definition
Addition of maps $f+g : A \to B$ is defined as the entry $h$ in the matrix product $\pmatrix{1_{AA}&f\\0_{BA}&1_{BB}} \cdot \pmatrix{1_{AA}&g\\0_{BA}&1_{BB}} = \pmatrix{1_{AA}&h\\0_{BA}&1_{BB}}$, where $f,g : A \to B$.
The exercise by Lawvere in this section asks to prove that matrix multiplication 'works' in linear categories. 
Problem is, even after having received an answer to my previous question, I am still at a complete loss as to how one might prove this. How does one prove this identity?
 A: (Note: I write compositions from left to right.)


*

*Since we know $A+B\cong A\times B$ and that products and coproducts are only determined up to isomorphisms, we can eventually choose the same object for both. 
More specifically, supposed we are given a choice of product and coproduct objects, together with their projections and inclusions, we can adjust e.g. the inclusions and use the chosen product object $A\times B$ to serve as coproduct. 
By the conditions, it follows that the maps arising by the product rule $(1,\,0):A\to A\times B$ and $(0,\,1):B\to A\times B$ will be the corresponding injections, and then $\pmatrix{1_A\\0_{AB}}=\pi_A$ and $\pmatrix{0_{AB}\\1_B}=\pi_B$. 
With this setting, $\pmatrix{1_A&0_{AB}\\0_{BA}&1_B}=1_{A\times B}=\alpha$.

*Obviously, the matrix entries can be recovered by composing with the adequate inclusions and projections. 
So, on one hand, it's enough to prove the base case 
$$\pmatrix{f_{AX}& f_{AY}}\pmatrix{g_{XU}\\g_{YU}}\ =\ 
f_{AX}\,g_{XU}\,+\,f_{AY}\,g_{YU}$$
on the other hand, for $f,g:A\to B$, the definition of $f+g$ boils down to
$$f+g\ :=\ \pmatrix{1_A& f}\pmatrix{g\\1_B}$$
Well, this is not the lucky order of addition for our purpose (and we haven't yet proved commutativity). But, applying the natural swap $A\times B\to B\times A$, we arrive to 
$$\pmatrix{1_A& f}\pmatrix{g\\1_B}\ =\  \pmatrix{f&1_A}\pmatrix{1_B\\g}$$

*
Put it together, we want to prove
  $$\pmatrix{a&b}\pmatrix{c\\d}=\pmatrix{ac&1}\pmatrix{1\\bd}$$
  provided $a:A\to X,\ b:A\to Y,\ c:X\to U,\ d:Y\to U$.


*This can be done in two steps (which are dual to each other):
$$\underset{A\ \to\ X\times Y\ \to\ U}{\pmatrix{a&b}\pmatrix{c\\d}}\ =\ \underset{A\ \to\ U\times Y\ \to\ U}{\pmatrix{ac&b}\pmatrix{1_U\\d}}\ =\ \underset{A\ \to\ U\times A\ \to\ U}{\pmatrix{ac&1_A}\pmatrix{1_U\\bd}}$$
For the first equation, consider the map $\vartheta:X\times Y\to U\times Y$, induced by $c:X\to U$ and $1_Y$ according to the product structures. 
(Observe that effectively $\vartheta=\pmatrix{c&0\\0&1_Y}$, and consequently it coincides with the map induced by $c$ and $1_Y$ for the coproduct structures.) 
Then verify that we have $\pmatrix{a&b}\,\vartheta=\pmatrix{ac&b}$ and $\vartheta\,\pmatrix{1_U\\d}=\pmatrix{c\\d}$, so that it's just the two associations of the triple composition of morphisms
$$\pmatrix{a&b}\pmatrix{c&0\\0&1}\pmatrix{1\\d}$$
The other equation can be proven similarly.

